The class width of the frequency distribution and tell whether the original data values can be identified.

Question

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Answer 1

Question

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

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Answer 2

Question

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

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Answer 3

Question

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

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Answer 4

Question

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

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Answer 5

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

Answer 6

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

Answer 7

Chapter 2, Problem 1CQQ

To determine

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Answer 8

Expert Solution & Answer

Answer to Problem 1CQQ

The class width is 20.

No, the original data values cannot be identified.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The frequency distribution table with the following class limits is provided.

Classes	Frequency
0-19	3
20-39	18
40-59	21
60-79	15
80-99	6
100-119	5
120-139	0
140-159	1

Concept used:

Class width is defined as the difference between two consecutive lower class limits in a frequency distribution. The formula to calculate the class width is given:

Class width = Second lower class limit−First lower class limit

The above formula is true for all the given lower class limits.

Calculation:

The total number of classes is 8. The first and second lower class limits are 0 and 20. Calculate the class width as follows:

Class width = Second lower class limit−First lower class limit=20−0=20

Hence, the class width is 20.

It is not possible to identify all the original data values since the data in the frequency distribution is provided in terms of intervals and the corresponding data values in those intervals are given in terms of frequency.

Thus, it is not possible to identify the original data values.

Answer 12

Ch. 2.1 - Prob. 1BSC Ch. 2.1 - Prob. 2BSC Ch. 2.1 - Prob. 3BSC Ch. 2.1 - Prob. 4BSC Ch. 2.1 - Prob. 5BSC Ch. 2.1 - Prob. 6BSC Ch. 2.1 - Prob. 7BSC Ch. 2.1 - Prob. 8BSC Ch. 2.1 - Prob. 9BSC Ch. 2.1 - Prob. 10BSC

The class width of the frequency distribution and tell whether the original data values can be identified.

To find: The class width of the frequency distribution and tell whether the original data values can be identified.

Answer to Problem 1CQQ

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions